Answer
$\dfrac{5}{13}$
Work Step by Step
Let us substitute
$g(h^{-1} (\dfrac{12}{5}))=\cos [\tan^{-1}(\dfrac{12}{5})]$
Suppose that $\theta=\tan^{-1} (\dfrac{12}{5})$
The range of $\tan^{-1}$ is $\frac{-\pi}{2} \lt \theta \lt \frac{\pi}{2}$ (quadrants I and IV). Since $\frac{12}{5}$ is positive, $\theta$ is in quadrant I.
Because we know the opposite and the side adjacent to the angle, it makes sense for us to use the tangent function:
$\tan {\theta}=\dfrac{opposite}{Adjacent}=\dfrac{12}{5}$
We wish to find the cosine function, which can be written as: $\cos (\theta)=\dfrac{Adjacent}{hypotenuse}$. So, we need to take the help of the Pythagorean Theorem for a right triangle to find the hypotenuse
$=\sqrt{(12)^2+(5)^2}=\sqrt {169}=13$
Note that we used positive $5$ and positive $13$ because $\theta$ is in quadrant I.
Since, $\cos (\theta)=\dfrac{Adjacent}{hypotenuse}=\dfrac{5}{13}$
Thus, $\cos [\tan^{-1}(\dfrac{12}{5})]=\dfrac{5}{13}$
